How do you solve #log_2 (x+2) - log_2 (x-5) = 3#?

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Answer 1 · 2016-01-02T15:48:10

Unify the logarithms and cancel them out with #log_(2)2^3#

#x=6#

#log_(2)(x+2)+log_(2)(x-5)=3#

Property #loga-logb=log(a/b)#

#log_(2)((x+2)/(x-5))=3#

Property #a=log_(b)a^b#

#log_(2)((x+2)/(x-5))=log_(2)2^3#

Since #log_x# is a 1-1 function for #x>0# and #x!=1#, the logarithms can be ruled out:

#(x+2)/(x-5)=2^3#

#(x+2)/(x-5)=8#

#x+2=8(x-5)#

#x+2=8x-8*5#

#7x=42#

#x=42/7#

#x=6#